October 6, 2013
Tongjia Shi
Moments of the Length of k th Shortest Cycle
1
1.1
Random z-permutation as Unit Poisson Process
Like for the longest cycles, consider a unit poisson process on the real line.
Let
j−1 k
X
z
tj (z) = θ
,
k = 1, 2, 3, · · · .
k
k=1
and
t∞ (z) = lim tj (z) = θ log
j→∞
1
.
1−z
Then there is a natural coupling between the unit poisson process on [0, t∞ (z)]
and the (θ-biased) random z-permutation. If there are q jumps in the j th
interval then draw q cycles with length j. The probability density at t of
the rth jump of the Poisson process on [t0 , t∞ (z)] is
e−t
tr−1
,
(r − 1)!
−∞ < t ≤ t∞ (z).
The mth moment of the length of the rth shortest cycle is then
Z tj+1 (z)
∞
X
tr−1
m
m
Ez [(Sr ) ] =
j
e−t
dt.
(r − 1)!
tj (z)
j=1
1.2
Change of variable
Make a change of variable
∞
Z
t∞ (z) − t = θE(x) =
x
Let z =
e−s ,
e−y
dy.
y
0 < s < ∞. Since
t∞ (e−s ) − tj (e−s ) = θ
∞ −ks
X
e
k=j
and
E(js) <
∞ −ks
X
e
k=j
k
j
,
< E((j − 1)s),
1
j = 1, 2, · · ·
j = 1, 2, · · · ,
there exists xj (s) such that (j − 1)s < xj (s) < js and
∞ −ks
X
e
= E(xj (s)).
k
k=j
So we have
Ez [(Sr )m ] =
∞
X
jm
Z
xj+1 (s)
xj (s)
j=1
= (1 − z)
θ
∞
X
j
m
Z
xj+1 (s)
xj (s)
j=1
1.3
eθE(x)−t∞
[t∞ − θE(x)]r−1 θe−x
dx
(r − 1)!
x
[t∞ − θE(x)]r−1 θeθE(x)−x
dx
(r − 1)!
x
Derivation Using Tauberian Theorem
Notice that
∞
∞
X
X
[xj (s)]m µj < Ez [(Sr )m ] <
[xj+1 (s)]m µj ,
j=1
j=1
with
Z
xj+1 (s)
µj =
e−θE(x)
xj (s)
E(x)r−1 θe−x
dx.
(r − 1)! x
Assuming m are θ are not both one. This is an approximation of the Riemann integral as s → 0. So s/(1 − z) → 1, x1 (s) → 0 and
(1 − z)m θ
lim
Ez [(Sr )m ] = (1 − z)θ ln
z→1
m!
where
θ
Hr,m
θ
=
(r − 1)!
Z
∞
0
1
1−z
r−1
θ
Hr,m
xm−1 θE(x)−x
e
dx.
m!
We rewrite this as
(1 − z)m−θ
ln
lim
z→1
m!
1
1−z
−(r−1)
2
θ
Eθz [(Sr )m ] = Hr,m
We will now use Tauberian Theorems. First,
−(r−1)
(1 − z)m−θ
1
θ
Hr,m = lim
ln
Eθz [(Sr )m ]
z→1
m!
1−z
−(r−1) X
∞
1
Γ(n + θ)
(1 − z)m−θ
ln
(1 − z)θ z n En [(Sr )m ]
= lim
z→1
m!
1−z
n!Γ(θ)
n=0
−(r−1)
∞
X
Γ(n + θ)
1
m
m
= lim
zn
En [(Sr ) ](1 − z) ln
z→1
m!n!Γ(θ)
1−z
n=0
Since the coefficients of z n are nonnegative, we have
n
X
Γ(n + θ)
θ
En [(Sr )m ] ∼ Γ(m + 1)−1 nm ln(n)r−1 Hr,m
,
m!n!Γ(θ)
n → ∞.
k=0
Now, we will show that
Γ(n + θ)
En [(Sr )m ] is nondecreasing in n. This is
m!n!Γ(θ)
equivalent to
(n + 1)En [Srm ] ≤ (n + θ)En+1 [Srm ].
Notice that
En [Srm ] = En+1 [En [Srm ]]
Sr
n + θ + 1 − Sr m
m
≤ En+1
(Sr − 1) −
Sr
n+θ+1
n+θ+1
mEn+1 [Srm ]
≤ En+1 [Srm ] +
n+θ+1
So we only need
1−θ
mEn+1 [Srm ]
≤
,
n+θ
(n + θ + 1)En+1 [Srm ]
which is true for sufficiently large n. Hence
Γ(n + θ)
θ
En [(Sr )m ] ∼ Γ(m+1)−1 [mnm−1 ln(n)r−1 +nm−1 ln(n)r−2 ]Hr,m
,
m!n!Γ(θ)
m nθ
Sr
θ
En
∼ mΓ(θ)Hr,m
,
n → ∞.
r−1
(ln n)
n
In fact the right hand side is a constant, so
m Z ∞ m−1
nθ
θ
Sr
x
lim
En
= mΓ(θ)
eθE(x)−x dx.
n→∞ (ln n)r−1
n
(r − 1)! 0
m!
3
n → ∞.
Rewriting this we get
nθ
En
n→∞ (ln n)r−1
lim
1.4
Sr
n
m =
Γ(1 + θ)
(r − 1)!
Z
0
∞
xm−1 θE(x)−x
e
dx.
(m − 1)!
Special Cases
When θ = 1 and m 6= 1. The mth moment of the rth shortest cycle (in a
unbiased random permutation) satisfies
m Z ∞
Sr
1
xm−1 E(x)−x
n
=
E
e
dx.
lim
n
n→∞ (ln n)r−1
n
(r − 1)! 0 (m − 1)!
agreeing with Shepp and Lloyd’s result.
When θ = 1/2. The expected length of the shortest cycle is (m = 1, r =
1)
√ Z ∞
√
Sr
π
lim nEn
=
eE(x)/2−x dx.
n→∞
n
2 0
agreeing with Pippenger’s result.
4